Error in rounding off long double

[unkn00wn]

So, i wrote this code to round off long double. But its showing some error beyond me.

long double round(long double a,int i)
{
long double temp=0;
long long ii=0;
ii=pow(10,i);

temp=a*ii;
cout<<fixed<<setprecision(20)<<temp<<endl;
temp+=0.5;
cout<<temp<<endl;
temp=floor(temp);
cout<<temp<<endl;
temp=temp/ii;
cout<<temp<<endl;

return(temp);
}

203214164615.76216505467891693115
203214164616.26216505467891693115
203214164616.00000000000000000000
20.32141598719999999942

Any help!?
Why diving by long long giving approximation? HOw to fix this?

[Peter87]

What values do you pass to round?

[unkn00wn]

long double a=0,b=0;
a=20.32141646157621672176152;
cout<<fixed<<setprecision(20)<<"Round of "<<a<<" is "<<round(a,10)<<endl;

Also I tried

temp*=pow(10,-i)

It gives

203214164615.76216505467891693115
203214164616.26216505467891693115
203214164616.00000000000000000000000000000000000000000000000000
20.32141646160001127903449447131833949242718517780304
Round of 20.32141646157621650559 is 20.32141646160001127903

[Peter87]

Ah, ok, so you want to round to a certain decimal place. It's not really a good idea to do that because many floating point values can't be stored exactly so you will get rounding errors. If you just want to show a certain number of decimals it's much better to use setprecision when printing the value.

[unkn00wn]

Thank you for your reply.

As this is my assignment, i must have to do so!
I understand precision is only helpful in displaying the variable. But while doing operation on it it is using its actual value.

Any other way to do so?